Strange Question With Square Roots

I'm having a bit of a problem with this question. I've tried solving it but when I plug it back into the equation, it doesn't work. I've also come up with several different solutions so I'm hoping that someone can put up a solution to this. Here's the question:
[IMG]http://img140.imageshack.us/my.php?image=untitledlt1.png[/IMG

Thus,
Square,
Thus,
Thus,
Thus,
That tells us that the solution must be one of these two.
Substituting into the equation to check the "only if" condition we see that the negative solution doth not work.
Thus,
Is the only solution.

If you'd be interested to know where the "false solution" came from, it was introduced in the step where ThePerfectHacker took the square of both sides.

Example: x = 3. Of course, this 'equation' is already 'solved', but suppose I square both sides: x² = 9. Now, I have introduced the 'solution' x = -3, which isn't a solution to the initial equation.

If you're a bit uncomfortable with just squaring, possibly introducing false solutions, and in the end checking your answers, you can do the following: squaring is allowed if both sides have the same sign. Now in that step, we had:

But you know that (the double of) a square root is nonnegative, so the right hand side can't be negative either. This yields the condition: . If you take along this condition, all your steps remain equivalent equations and you'll see that the second solution is smaller than -1, hence you have to drop it.

Somtimes this is more work than just checking your answers, sometimes it isn't. In any case, it's 'elegant' in the way that you keep working with equivalent equation, not introducing any solutions because of the acquired condition on x.

I hope this at least 'clarifies' what's happenning here, mathemtically

If you'd be interested to know where the "false solution" came from, it was introduced in the step where ThePerfectHacker took the square of both sides.

Example: x = 3. Of course, this 'equation' is already 'solved', but suppose I square both sides: x² = 9. Now, I have introduced the 'solution' x = -3, which isn't a solution to the initial equation.

If you're a bit uncomfortable with just squaring, possibly introducing false solutions, and in the end checking your answers, you can do the following: squaring is allowed if both sides have the same sign. Now in that step, we had:

But you know that (the double of) a square root is nonnegative, so the right hand side can't be negative either. This yields the condition: . If you take along this condition, all your steps remain equivalent equations and you'll see that the second solution is smaller than -1, hence you have to drop it.

Somtimes this is more work than just checking your answers, sometimes it isn't. In any case, it's 'elegant' in the way that you keep working with equivalent equation, not introducing any solutions because of the acquired condition on x.

I hope this at least 'clarifies' what's happenning here, mathemtically

I have a different explanation why a "false" solution appears. This is how I think of it.

In a strictly formal level, when we solve an equation we have a chain of conditional statements. For example,
Then,
This is a conditional statement.
Now, there is a logical error mistake by assuming the converse it true.
For example,
If,
Then,
Then,
Since conclusion is correct then hypothesis (2=6) was correct... wrong!
In this case the converse is not true.
In symbolic logic,
Meaning we cannot conclude what the hypothesis is.
If you remember Soroban's parametric circle riddle, just this error, by assuming the converse.

That tells us when we solve the equation,
If,
Then,
We need to see if it is a solution, because the logical implication tells us, that if a solution exists it must be this. And now we see whether or not it is true.
However, it is not necessary to check the solution, because,
If and only if,
Thus, whatever we get in the end as a solution must be a solution because of a biconditional statement.

Most algebraic equations involve biconditional statements. Meaning that when you solve it you get a solutions, and those are all the solutions and all of them work.

When dealing with a square root, a biconditional statment fails. For example,
If and only if (...wrong),
Because the statement other way around (converse) is not true.
We need to be especially careful in the structure of biconditional statements when we solve this equation.
If we do it, we will see that there is a step in the solution where the biconditional statement fails. Thus, the final solutions not necessarily work. However, those are the only possibilities.

Although I don't disagree with your explanation, I personally believe that this formal logic argument is beyond the scope of the avarage high school classes covering equation solving. The rule I gave for squaring the sides of an equation is rather easy, I think, and also applies to your example. In sqrt(x) = -4, squaring is only allowed if both sides have the same sign, but with the square root having to be nonnegative and -4 obviously negative, you cannot square - there is no solution (which was clear from inspection).

When solving these irrational equations, I usually tell students to check two conditions: the "squaring condition" and the "existence condition". As I said, the "squaring condition" requires both sides to have the same sign if you want to square. The "existence condition" requires that expression under square roots are nonnegative, in order to have real solutions (I'm talking about solving in IR of course).